Problem: $ g(x) = \int_{-10}^x\sqrt{t^2 + 11}\,dt\,$ $ g'(-5)\, =$
Solution: The Fundamental Theorem of Calculus If $~ g(x)=\int_a^xf(t)\,dt\,$, then $~g^\prime (x)=f(x)\,$ This only works if $f$ is continuous on $[a,b]$. Thankfully, the function $f(t) = \sqrt{t^2 + 11}$ is continuous on $[-10,-5]$. Applying the theorem We're given: $ g(x) = \int_{-10}^x\sqrt{t^2 + 11}\,dt\,$ So the theorem tells us: $ g\,^\prime(x) =\sqrt{x^2 + 11}$ Evaluating $g'(-5)$ $ g'(-5) = \sqrt{(-5)^2 + 11}=\sqrt{36}=6$ The answer: $g'(-5)=6$